# Let R be the region enclosed by f(x) = sinx, g(x) =1-x, and x=0. What is the volume of the solid produced by revolving R around the x-axis?

Aug 9, 2017

#### Explanation:

Here is a picture of the region with a slice taken perpendicular to the axis of rotation.

Let $c =$ the point of intersection of $\sin x$ and $1 - x$.

The volume of the solid is

$V = \pi {\int}_{0}^{c} \left({\left(1 - x\right)}^{2} - {\sin}^{2} x\right) \mathrm{dx}$

 = pi int_0^c ((1-x)^2 - 1/2(1-cos(2x)) dx

$= \pi {\left[- {\left(1 - x\right)}^{3} / 3 - \frac{1}{2} x + \frac{1}{2} \sin x \cos x\right]}_{0}^{c}$

$= \pi \left(- {\left(1 - c\right)}^{3} / 3 + \frac{c}{2} + \frac{1}{2} \sin \left(c\right) \cos \left(c\right) + \frac{1}{3}\right)$

If desired, we can rewrite using

$\sin c = 1 - c$ and $\cos c = \sqrt{2 c - {c}^{2}}$.

Or we can evaluate using $c \approx 0.51$